05 August 2008
Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Example:
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Input: nums = [1,2,3]
Output:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
同样一道回溯法的题目，解法和分析一模一样。
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class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> results = new ArrayList<>();
if (nums == null || nums.length == 0) {
return results;
}
Arrays.sort(nums);
backtracking(nums, results, new ArrayList<Integer>(), 0);
return results;
}
private void backtracking(int[] nums, List<List<Integer>> results, List<Integer> oneResult, int index) {
results.add(new ArrayList<>(oneResult)); // 把add放在这里，也可以放在循环里面，后者需要事先单独加上题目要求的一个空结果
for (int i = index; i < nums.length; i++) {
if (oneResult.contains(nums[i])) { //重复元素
continue;
}
oneResult.add(nums[i]);
backtracking(nums, results, oneResult, i + 1); // 注意这里是i + 1而不是index + 1
oneResult.remove(oneResult.size() - 1);
}
}
}
结果添加在for循环之中，事先加好空的第一个结果
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class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> results = new ArrayList<>();
List<Integer> oneResult = new ArrayList<>();
Arrays.sort(nums); // 因为没有重复，不排序也会遍历所有组合，排序会有些小优化路径
results.add(oneResult);
dfs(nums, results, oneResult, 0);
return results;
}
private void dfs(int[] nums, List<List<Integer>> results, List<Integer> oneResult, int index) {
for (int i = index; i < nums.length; i++) {
if (oneResult.contains(nums[i])) {
continue;
}
oneResult.add(nums[i]);
results.add(new ArrayList<>(oneResult));
dfs(nums, results, oneResult, i + 1); // i + 1表示将当前元素的后面所有元素加上
oneResult.remove(oneResult.size() - 1);
}
}
}